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Fredholm equation of the second kind

  1. Oct 16, 2008 #1
    1. The problem statement, all variables and given/known data

    I need to prove that a Fredholm equation of the second kind is functionally linear.

    I'm not sure what functional linearity is or if that's exactly what is being asked because it's not in my numerical analysis book and everytime i look on the internet I'm referred to linear algebra.

    Can someone get me on the right track as far as what i'm actually looking to prove?

    edit: I also need to solve a fredholm eqn of the second kind analytically. Hint: my result should contain a low order polynomial.

    it is y(x) = x + integral[0,1](x^2*t)y(t)dt

    every example i see is numerical integration. so i couldn't find any examples to work off of.
    should i do integration by parts on the integrand and then set up a differential equation involving Y(t) and y(x) and then solve with a characteristic polynomial?
  2. jcsd
  3. Oct 17, 2008 #2


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    Saying that an operator, U(f), is "functionally linear" means that U(af+ bg)= aU(f)+ bU(g) where f and g are functions and a and b are numbers.

    The equation you are asked to solve is [itex]y(x)= x+ \int_0^1 x^2ty(t)dt= x+ x^2\int_0^1 ty(t)dt[/itex].

    If you let [itex]A= \int_0^1 ty(t)dt[/itex], a number, then your equation is simply y(x)= x+ Ax2. that's the "low order polynomial" you mention. You only need to determine what A is. Multiply on both sides by x to get xy(x)= x2+ Ax3 and integrate from 0 to 1. The left side is just X so you get a linear equation to solve for x.
  4. Oct 20, 2008 #3
    thanks so much that helps everything.
  5. Oct 20, 2008 #4
    uh, apparently i need the proof in the form of c1y1+c2y2 = f and Lf=y. in this case do i need to use method of undetermined coeffecients with the y(x)=x+ax^2 ?
  6. Oct 20, 2008 #5
    bump. Any help is greatly appreciated!
  7. Oct 20, 2008 #6
    The Rock says...

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